From the shape of your curve, I guess that you might get roughly the shape you need by setting
yd = (y+fliplr(y))/2
where I'm assuming that your "raw" vector y is a row. Change fliplr into flipud if it is a column (or transpose y).
But then I'm not sure that the data you're getting would have any meaning.
Update: Actually I think that my solution gives something similar to the red curve, but the plateau would be below the baseline.
I took a look at the data you attach, which covers the part from the second to the third curve in your plot. Your vector is a column, so flipud should apply.
Perhaps what you are looking for is
yd = (y+flipud(y))/2;
b = yd(1);
yd = -(yd-b)+b;
In the last line I change the sign of yd after subtracting the baseline, so that the plateau is on top of the baseline (which now is 0) and add the baseline again.
This should give roughly the shape you're looking for, if y is the whole data you used in the plot (not only what you attach).
I have strong doubts about any data analysis based on yd, though.
